HW8Soln - Solutions to HW8 (due 4/2, 4/3) Page 274 7: dy dx...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to HW8 (due 4/2, 4/3) Page 274 7: dy dx = dy/dt dx/dt = e x + y y- 1 = e x e y y- 1 . This is separable, leading to Z ( y- 1) e- y dy = Z e x dx + C, or- ye- y = e x + C. or e x + ye- y = C The mixture of y and e- y means we cannot solve explicitly for y in terms of x , but, we can solve for x in terms of y , x = ln (- ye- y- C ) 8: dy dx = dy/dt dx/dt = 3 x 2- 2 xy x 2- 2 y- 3 This is not separable or linear, so we check to see if it is exact. First rewrite in the form (3 x 2- 2 xy ) dx + (- x 2 + 2 y- 3 ) dy = 0 If this is Mdx + Ndy , then M/y =- 2 x and N/x =- 2 x , so the equation is exact. To find F ( x, y ) such that F/x = M , we integrate with respect to x (holding y constant). We get that F ( x, y ) = x 3- x 2 y + f ( y ) for some function f ( y ) depending only on y . Taking the partial with respect to y and setting it equal to N we get- x 2 + f ( y ) =- x 2 + 2 y- 3 so that f ( y ) = 2 y- 3 and f ( y ) =- y- 2 + C . Therefore we can take F ( x, y ) = x 3- x 2 y- y- 2 so that the integral curves are of the form x 3- x 2 y- y- 2 = C There is no nice explicit form for this. 9: dy dx = dy/dt dx/dt = e x + y 2 y- x This is not separable or linear, so we check to see if it is exact. First rewrite in the form ( e x + y ) dx + ( x- 2 y ) dy = 0 . If this is Mdx + Ndy , then M/y = 1 and N/x = 1, so the equation is exact. To find F ( x, y ) such that F/x = M , we integrate with respect to x (holding y constant). We get that F ( x, y ) = e x + xy + f ( y ) for some function f ( y ) depending only on y . Taking the partial with respect to y and setting it equal to N we get x + f ( y ) = x- 2 y so that f ( y ) =- 2 y and f ( y ) =- y 2 + C . Therefore we can take F ( x, y ) = e x + xy- y 2 so that the integral curves are of the form e x + xy- y 2 = C . This can be solved for y using the quadratic formula, y = x p x 2 + 4( C- e x ) 2 , 1 but in most applications, that form is not more desirable than the implicit form. 10: To find critical points, solve x 2- 1 = 0 , xy = 0. The first equation implies x = 1. Combining that with the second equation we get that y = 0. Thus there are two critical points, (1 , 0), and (- 1 , 0). The phase plane equation is dy dx = dy/dt dx/dt = xy x 2- 1 which is separable leading to Z dy y = Z x dx x 2- 1 + C....
View Full Document

Page1 / 5

HW8Soln - Solutions to HW8 (due 4/2, 4/3) Page 274 7: dy dx...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online